Let $A, B$ be commutative unital Banach algebras and let $\varphi: A \rightarrow B$ be a continuous unital map such that $$\overline{\varphi(A)} = B$$

Let $$\varphi^{*}: \text{Max}(B) \rightarrow \text{Max}(A)$$ $$\varphi^{*}(m) = m(\varphi)$$ be the map from the space of maximal ideals of $B$ to the space of maximal ideals of $A$ induced by $\varphi$.

How to prove that $\varphi^{*}$ is a topologically injective map?

(Recall that an operator $T: X \rightarrow Y$ is called topologically injective if $T: X \rightarrow \text{im}(T)$ is a homeomorphism)

My progress on the problem is the following: first of all, the space of maximal ideals of a commutative Banach algebra $A$ can be identified with the space of continuous functionals of the form $m: A \rightarrow \mathbb{C}$. Clearly the map above is continuous, since the pointwise convergence of a net $(n_{i})$ in $\text{Max}(B)$ implies the convergence of the net $ m_{i} \circ \varphi) $

(the space of continuous linear functional is endowed with the weak* topology)

If we assume for the moment that the map is bijective, then the fact that a continuous bijective map between compact Hausdorff spaces is a homeomorphism, yields the result.

(here the space of maximal ideals is compact in weak* topology since the algebra is unital)

The suggested proposition looks like a relaxation of the aformentioned reasoning above though i cannot figure out an easy way to modify it to make it work. Are there any hints?

  • $\begingroup$ You need to use the fact that all characters in the spectrum are of norm one to relate the topology of $\mathrm{im}(\varphi^\ast)$ with the topology of $\mathrm{Max}(B)$. $\endgroup$ – Adrián González-Pérez Dec 26 '18 at 13:45

Use that all characters are of norm one.

Indeed, they are bounded since their kernel is a maximal ideal and therefore closed. The fact that they are contractive follows from spectral theory. If $a - \lambda 1$ is invertible so is $\chi(a) - \lambda$. The counter-reciprocal gives that when $\lambda$ is in the image of $a$ under $\chi$ then $a - \lambda 1$ is not invertible and so $$ |\chi(a)| \leq \sup \{ |\lambda| : \lambda \in \mathrm{sp}(a) \} \leq \| a \| $$

Using the fact that $\varphi[A]$ is dense in $B$ you have that if two continuous functional $\varphi_1$ and $\varphi_2$ agree on $\varphi[A]$, then they are equal. This gives the injectivity.

For topological injectivity you need to see that if $\varphi^\ast(\chi_n) = \chi_n \circ \varphi \in \mathrm{im}(\varphi^*)\subset \mathrm{Max}(A)$ converge to $\chi \circ \varphi$ then $\chi_n \to \chi$ in $\mathrm{Max}(B)$. Let $b \in B$, we can find $a_\epsilon$ with $\|\varphi(a_\epsilon) - b\| < \epsilon$ and \begin{eqnarray*} |\chi_n(b) - \chi(b) | & \leq &\| \chi_n(b) - \chi_n(\varphi(a_\epsilon))\| + |\chi_n(\varphi(a_\epsilon)) - \chi(\varphi(a_\epsilon))| + | \chi(\varphi(a_\epsilon)) - \chi(b)|\\ & \leq & \epsilon + |\chi_n(\varphi(a_\epsilon)) - \chi(\varphi(a_\epsilon))| + \epsilon. \end{eqnarray*} But that implies that the limit of $|\chi_n(b) - \chi(b)|$ is smaller or equal than $\epsilon$ for every $\epsilon$ and therefore $0$.


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